NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Straight Lines

Question1: (a) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the x-axis.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of x-axis is $y=0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m' = 0$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m = \frac{k-3}{4-k^2}$
Now,
$m=m'$
$\frac{k-3}{4-k^2}=0$
$k-3=0$
$k=3$
Therefore, the value of k is 3

Question1: (b) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the y-axis.

Answer:

Given the equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of y-axis is $x = 0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m' = \infty = \frac{1}{0}$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m = \frac{k-3}{4-k^2}$
Now,
$m=m'$
$\frac{k-3}{4-k^2}=\frac{1}{0}$
$4-k^2=0$
$k=\pm2$
Therefore, value of k is $\pm2$

Question1: (c) Find the values of k for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is Passing through the origin.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
It is given that it passes through origin (0,0)
Therefore,
$(k-3).0-(4-k^2).0+k^2-7k+6=0$
$k^2-7k+6=0$
$k^2-6k-k+6=0$
$(k-6)(k-1)=0$
$k = 6 \ or \ 1$
Therefore, value of k is $6 \ or \ 1$

Question 2: Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $\small 1$ and $\small -6$, respectively.

Answer:

Let the intercepts on x and y-axis are a and b respectively
It is given that
$a+b = 1 \ \ and \ \ a.b = -6$
$a= 1-b$
$\Rightarrow b.(1-b)=-6$
$\Rightarrow b-b^2=-6$
$\Rightarrow b^2-b-6=0$
$\Rightarrow b^2-3b+2b-6=0$
$\Rightarrow (b+2)(b-3)=0$
$\Rightarrow b = -2 \ and \ 3$
Now, when $b=-2\Rightarrow a=3$
and when $b=3\Rightarrow a=-2$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}=1$

Case (i) when a = 3 and b = -2
$\frac{x}{3}+\frac{y}{-2}=1$
$\Rightarrow 2x-3y=6$

Case (ii) when a = -2 and b = 3
$\frac{x}{-2}+\frac{y}{3}=1$
$\Rightarrow -3x+2y=6$
Therefore, equations of lines are $2x-3y=6 \ and \ -3x+2y=6$

Question 3: What are the points on the $\small y$-axis whose distance from the line $\small \frac{x}{3}+\frac{y}{4}=1$ is $\small 4$ units.

Answer:

Given the equation of the line is
$\small \frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y=12$
Let's take point on y-axis is $(0,y)$
It is given that the distance of the point $(0,y)$ from line $4x+3y=12$ is 4 units
Now,
$d= \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$
In this problem $A = 4 , B=3 , C =-12 ,d=4\ \ and \ \ (x_1,y_1) = (0,y)$
$4 = \left | \frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}} \right |=\left | \frac{3y-12}{\sqrt{16+9}} \right |=\left | \frac{3y-12}{5} \right |$

Case (i)

$4 = \frac{3y-12}{5}$
$20=3y-12$
$y = \frac{32}{3}$
Therefore, the point is $\left ( 0,\frac{32}{3} \right )$ -(i)

Case (ii)

$4=-\left ( \frac{3y-12}{5} \right )$
$20=-3y+12$
$y = -\frac{8}{3}$
Therefore, the point is $\left ( 0,-\frac{8}{3} \right )$ -(ii)
Therefore, points on the $\small y$-axis whose distance from the line $\small \frac{x}{3}+\frac{y}{4}=1$ is $\small 4$ units are $\left ( 0,\frac{32}{3} \right )$ and $\left ( 0,-\frac{8}{3} \right )$

Question 4: Find perpendicular distance from the origin to the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \phi ,\sin \phi ).$.

Answer:

Equation of line passing through the points $(\cos \theta ,\sin \theta )$ and $(\cos \phi ,\sin \phi )$ is
$(y-\sin \theta )= \frac{\sin \phi -\sin \theta}{\cos \phi -\cos \theta}(x-\cos\theta)$
$\Rightarrow (\cos \phi -\cos \theta)(y-\sin \theta )= (\sin \phi -\sin \theta)(x-\cos\theta)$
$\Rightarrow y(\cos \phi -\cos \theta)-\sin \theta(\cos \phi -\cos \theta)=x (\sin \phi -\sin \theta)-\cos\theta(\sin \phi -\sin \theta)$
$\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\cos\theta(\sin \phi -\sin \theta)-\sin \theta(\cos \phi -\cos \theta)$$\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\sin(\theta-\phi)$
$(\because \cos a\sin b -\sin a\cos b = \sin(a-b) )$
Now, distance from origin(0,0) is
$d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right |$ $(\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)$
$d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |$
$d = \left | \cos\frac{\theta-\phi}{2} \right |$

Question 5: Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines $\small x-7y+5=0$ and $\small 3x+y=0$.

Answer:

Point of intersection of the lines $\small x-7y+5=0$ and $\small 3x+y=0$
$\left ( -\frac{5}{22},\frac{15}{22} \right )$
It is given that this line is parallel to y - axis i.e. $x=0$ which means their slopes are equal
Slope of $x=0$ is ,$m' = \infty = \frac{1}{0}$
Let the Slope of line passing through point $\left ( -\frac{5}{22},\frac{15}{22} \right )$ is m
Then,
$m=m'= \frac{1}{0}$
Now, equation of line passing through point $\left ( -\frac{5}{22},\frac{15}{22} \right )$ and with slope $\frac{1}{0}$ is
$(y-\frac{15}{22})= \frac{1}{0}(x+\frac{5}{22})$
$x = -\frac{5}{22}$
Therefore, equation of line is $x = -\frac{5}{22}$

Question 6: Find the equation of a line drawn perpendicular to the line $\small \frac{x}{4}+\frac{y}{6}=1$ through the point, where it meets the $\small y$-axis.

Answer:

given equation of line is
$\small \frac{x}{4}+\frac{y}{6}=1$
we can rewrite it as
$3x+2y=12$
Slope of line $3x+2y=12$ , $m' = -\frac{3}{2}$
Let the Slope of perpendicular line is m
$m = -\frac{1}{m'}= \frac{2}{3}$
Now, the ponit of intersection of $3x+2y=12$ and $x =0$ is $(0,6)$
Equation of line passing through point $(0,6)$ and with slope $\frac{2}{3}$ is
$(y-6)= \frac{2}{3}(x-0)$
$3(y-6)= 2x$
$2x-3y+18=0$
Therefore, equation of line is $2x-3y+18=0$

Question 7: Find the area of the triangle formed by the lines $\small y-x=0,x+y=0$ and $\small x-k=0$.

Answer:

Given equations of lines are
$y-x=0 \ \ \ \ \ \ \ \ \ \ \ -(i)$
$x+y=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
$x-k=0 \ \ \ \ \ \ \ \ \ \ \ -(iii)$
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are $(0,0), (k,-k) \ and \ (k,k)$
Now, we know that area of triangle whose vertices are $(x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3)$ is
$A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$A= \frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|$
$A= \frac{1}{2}|k^2+k^2|$
$A= \frac{1}{2}|2k^2|$
$A= k^2$
Therefore, area of triangle is $k^2 \ square \ units$

Question 8: Find the value of $\small p$ so that the three lines $\small 3x+y-2=0,px+2y-3=0$ and $\small 2x-y-3=0$ may intersect at one point.

Answer:

Point of intersection of lines $\small 3x+y-2=0$ and $\small 2x-y-3=0$ is $(1,-1)$
Now, $(1,-1)$ must satisfy equation $px+2y-3=0$
Therefore,
$p(1)+2(-1)-3=0$
$p-2-3=0$
$p=5$
Therefore, the value of p is $5$

Question 9: If three lines whose equations are $y=m_1x+c_1,y=m_2x+c_2$ and $y=m_3x+c_3$ are concurrent, then show that $m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$.

Answer:

Concurrent lines means they all intersect at the same point
Now, given equation of lines are
$y=m_1x+c_1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
$y=m_2x+c_2 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
$y=m_3x+c_3 \ \ \ \ \ \ \ \ \ \ \ -(iii)$
Point of intersection of equation (i) and (ii) $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$

Now, lines are concurrent which means point $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$ also satisfy equation (iii)
Therefore,

$\frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3.\left ( \frac{c_2-c_1}{m_1-m_2} \right )+c_3$

$m_1c_2-m_2c_1= m_3(c_2-c_1)+c_3(m_1-m_2)$

$m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

Hence proved

Question 10: Find the equation of the lines through the point $\small (3,2)$ which make an angle of $\small 45^{\circ}$ with the line $\small x-2y=3$.

Answer:

Given the equation of the line is
$\small x-2y=3$
The slope of line $\small x-2y=3$ , $m_2= \frac{1}{2}$
Let the slope of the other line is, $m_1=m$
Now, it is given that both the lines make an angle $\small 45^{\circ}$ with each other
Therefore,
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan 45^\circ = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |$
$1= \left | \frac{1-2m}{2+m} \right |$
Now,

Case (i)
$1=\frac{1-2m}{2+m}$
$2+m=1-2m$
$m = -\frac{1}{3}$
Equation of line passing through the point $\small (3,2)$ and with slope $-\frac{1}{3}$
$(y-2)=-\frac{1}{3}(x-3)$
$3(y-2)=-1(x-3)$
$x+3y=9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Case (ii)
$1=-\left ( \frac{1-2m}{2+m} \right )$
$2+m=-(1-2m)$
$m= 3$
Equation of line passing through the point $\small (3,2)$ and with slope 3 is
$(y-2)=3(x-3)$
$3x-y=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$

Therefore, equations of lines are $3x-y=7$ and $x+3y=9$

Question 11: Find the equation of the line passing through the point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.

Answer:

Point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ is $\left ( \frac{1}{13},\frac{5}{13} \right )$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}= 1$
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
$x+y = a$ -(i)
It passes through point $\left ( \frac{1}{13},\frac{5}{13} \right )$
Therefore,
$a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}$
Put the value of a in equation (i)
we will get
$13x+13y=6$
Therefore, equation of line is $13x+13y=6$

Question 12: Show that the equation of the line passing through the origin and making an angle $\small \theta$ with the line $\small y=mx+c$ is $\small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$.

Answer:

Slope of line $\small y=mx+c$ is m
Let the slope of other line is m'
It is given that both the line makes an angle $\small \theta$ with each other
Therefore,
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan \theta = \left | \frac{m-m'}{1+mm'} \right |$
$\mp(1+mm')\tan \theta =(m-m')$
$\mp\tan \theta +m'(\mp m\tan\theta+1)= m$
$m'= \frac{m\pm \tan \theta}{1\mp m\tan \theta}$
Now, equation of line passing through origin (0,0) and with slope $\frac{m\pm \tan \theta}{1\mp m\tan \theta}$ is
$(y-0)=\frac{m\pm \tan \theta}{1\mp m\tan \theta}(x-0)$
$\frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}$
Hence proved

Question 13: In what ratio, the line joining $\small (-1,1)$ and $\small (5,7)$ is divided by the line $x+y=4$ ?

Answer:

Equation of line joining $\small (-1,1)$ and $\small (5,7)$ is
$(y-1)= \frac{7-1}{5+1}(x+1)$
$\Rightarrow (y-1)= \frac{6}{6}(x+1)$
$\Rightarrow (y-1)= 1(x+1)$
$\Rightarrow x-y+2=0$
Now, point of intersection of lines $x+y=4$ and $x-y+2=0$ is $(1,3)$
Now, let's suppose point $(1,3)$divides the line segment joining $\small (-1,1)$ and $\small (5,7)$ in $1:k$
Then,
$(1,3)= \left ( \frac{k(-1)+1(5)}{k+1},\frac{k(1)+1(7)}{k+1} \right )$
$1=\frac{-k+5}{k+1} \ \ and \ \ 3 = \frac{k+7}{k+1}$
$\Rightarrow k =2$
Therefore, the line joining $\small (-1,1)$ and $\small (5,7)$ is divided by the line $x+y=4$ in ratio $1:2$

Question 14: Find the distance of the line $\small 4x+7y+5=0$ from the point $\small (1,2)$ along the line $\small 2x-y=0$.

Answer:

point $\small (1,2)$ lies on line $2x-y =0$
Now, point of intersection of lines $2x-y =0$ and $\small 4x+7y+5=0$ is $\left ( -\frac{5}{18},-\frac{5}{9} \right )$
Now, we know that the distance between two point is given by
$d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
$d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|$
$d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|$
$d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |$
$d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}$
Therefore, the distance of the line $\small 4x+7y+5=0$ from the point $\small (1,2)$ along the line $\small 2x-y=0$ is $\frac{23\sqrt5}{18} \ units$

Question 15: Find the direction in which a straight line must be drawn through the point $\small (-1,2)$ so that its point of intersection with the line $\small x+y=4$ may be at a distance of $3$ units from this point.

Answer:

Let $(x_1,y_1)$ be the point of intersection
it lies on line $\small x+y=4$
Therefore,
$x_1+y_1=4 \\ x_1=4-y_1\ \ \ \ \ \ \ \ \ \ \ -(i)$
Distance of point $(x_1,y_1)$ from $\small (-1,2)$ is 3
Therefore,
$3= |\sqrt{(x_1+1)^2+(y_1-2)^2}|$
Square both the sides and put value from equation (i)
$9= (5-y_1)^2+(y_1-2)^2\\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1\\ 2y_1^2-14y_1+20=0\\ y_1^2-7y_1+10=0\\ y_1^2-5y_1-2y_1+10=0\\ (y_1-2)(y_1-5)=0\\ y_1=2 \ or \ y_1 = 5$
When $y_1 = 2 \Rightarrow x_1 = 2$ point is $(2,2)$
and
When $y_1 = 5 \Rightarrow x_1 = -1$ point is $(-1,5)$
Now, slope of line joining point $(2,2)$ and $\small (-1,2)$ is
$m = \frac{2-2}{-1-2}=0$
Therefore, line is parallel to x-axis -(i)

or
slope of line joining point $(-1,5)$ and $\small (-1,2)$
$m = \frac{5-2}{-1+2}=\infty$
Therefore, line is parallel to y-axis -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

Question 16: The hypotenuse of a right angled triangle has its ends at the points $\small (1,3)$ and $\small (-4,1)$ Find an equation of the legs (perpendicular sides) of the triangle.

Answer:

Slope of line OA and OB are negative times inverse of each other
Slope of line OA is , $m=\frac{3-y}{1-x}\Rightarrow (3-y)=m(1-x)$
Slope of line OB is , $-\frac{1}{m}= \frac{1-y}{-4-x}\Rightarrow (x+4)=m(1-y)$
Now,
for a given value of m we get these equations
If $m = \infty$
$1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0$
$x=1 \ \ \ \ and \ \ \ \ \ y =1$

Question 17: Find the image of the point $\small (3,8)$ with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

Answer:

Let point $(a,b)$ is the image of point $\small (3,8)$ w.r.t. to line $x+3y=7$
line $x+3y=7$ is perpendicular bisector of line joining points $\small (3,8)$ and $(a,b)$
Slope of line $x+3y=7$ , $m' = -\frac{1}{3}$
Slope of line joining points $\small (3,8)$ and $(a,b)$ is , $m = \frac{8-b}{3-a}$
Now,
$m = -\frac{1}{m'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
$\frac{8-b}{3-a}= 3$
$8-b=9-3a$
$3a-b=1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Point of intersection is the midpoint of line joining points $\small (3,8)$ and $(a,b)$
Therefore,
Point of intersection is $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$
Point $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$ also satisfy the line $x+3y=7$
Therefore,
$\frac{3+a}{2}+3.\frac{b+8}{2}=7$
$a+3b=-13 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$(a,b) = (-1,-4)$
Therefore, the image of the point $\small (3,8)$ with respect to the line $x+3y=7$ is $(-1,-4)$

Question 18: If the lines $\small y=3x+1$ and $\small 2y=x+3$ are equally inclined to the line $\small y=mx+4$ , find the value of $m$.

Answer:

Given equation of lines are
$\small y=3x+1 \ \ \ \ \ \ \ \ \ \ -(i)$
$\small 2y=x+3 \ \ \ \ \ \ \ \ \ \ -(ii)$
$\small y=mx+4 \ \ \ \ \ \ \ \ \ \ -(iii)$
Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
Slope of line $\small y=3x+1$ is , $\small m_1=3$
Slope of line $\small 2y=x+3$ is , $\small m_2= \frac{1}{2}$
Slope of line $\small y=mx+4$ is , $\small m_3=m$
Now, we know that
$\tan \theta = \left | \frac{m_1-m_2}{1+m_1m_2} \right |$
Now,
$\tan \theta_1 = \left | \frac{3-m}{1+3m} \right |$ and $\tan \theta_2 = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |$

It is given that $\tan \theta_1=\tan \theta_2$
Therefore,
$\left | \frac{3-m}{1+3m} \right |= \left | \frac{1-2m}{2+m} \right |$
$\frac{3-m}{1+3m}= \pm\left ( \frac{1-2m}{2+m} \right )$
Now, if $\frac{3-m}{1+3m}= \left ( \frac{1-2m}{2+m} \right )$
Then,
$(2+m)(3-m)=(1-2m)(1+3m)$
$6+m-m^2=1+m-6m^2$
$5m^2=-5$
$m= \sqrt{-1}$
Which is not possible
Now, if $\frac{3-m}{1+3m}= -\left ( \frac{1-2m}{2+m} \right )$
Then,
$(2+m)(3-m)=-(1-2m)(1+3m)$
$6+m-m^2=-1-m+6m^2$
$7m^2-2m-7=0$
$m = \frac{-(-2)\pm \sqrt{(-2)^2-4\times 7\times (-7)}}{2\times 7}= \frac{2\pm \sqrt{200}}{14}= \frac{1\pm5\sqrt2}{7}$

Therefore, the value of m is $\frac{1\pm5\sqrt2}{7}$

Question 19: If the sum of the perpendicular distances of a variable point $\small P(x,y)$ from the lines $\small x+y-5=0$ and $\small 3x-2y+7=0$ is always $\small 10$. Show that $\small P$ must move on a line.

Answer:

Given the equation of line are
$x+y-5=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
$3x-2y+7=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, perpendicular distances of a variable point $\small P(x,y)$ from the lines are

$d_1=\left | \frac{1.x+1.y-5}{\sqrt{1^2+1^2}} \right |$ $d_2=\left | \frac{3.x-2.y+7}{\sqrt{3^2+2^2}} \right |$
$d_1=\left | \frac{x+y-5}{\sqrt2} \right |$ $d_2=\left | \frac{3x-2y+7}{\sqrt{13}} \right |$
Now, it is given that
$d_1+d_2= 10$
Therefore,
$\frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10$
$(assuming \ x+y-5 > 0 \ and \ 3x-2y+7 >0)$
$(x+y-5)\sqrt{13}+(3x-2y+7)\sqrt2=10\sqrt{26}$

$x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt2$

Which is the equation of the line
Hence proved

Question 20: Find equation of the line which is equidistant from parallel lines $9x+6y-7=0$ and $3x+2y+6=0$.

Answer:

Let's take the point $p(a,b)$ which is equidistance from parallel lines $9x+6y-7=0$ and $3x+2y+6=0$
Therefore,
$d_1= \left | \frac{9.a+6.b-7}{\sqrt{9^2+6^2}} \right |$ $d_2= \left | \frac{3.a+2.b+6}{\sqrt{3^2+2^2}} \right |$
$d_1= \left | \frac{9a+6b-7}{\sqrt{117}} \right |$ $d_2= \left | \frac{3a+2b+6}{\sqrt{13}} \right |$
It is that $d_1=d_2$
Therefore,
$\left | \frac{9a+6b-7}{3\sqrt{13}} \right |= \left | \frac{3a+2b+6}{\sqrt{13}} \right |$
$(9a+6b-7)=\pm 3(3a+2b+6)$
Now, case (i)
$(9a+6b-7)= 3(3a+2b+6)$
$25=0$
Therefore, this case is not possible

Case (ii)
$(9a+6b-7)= -3(3a+2b+6)$
$18a+12b+11=0$

Therefore, the required equation of the line is $18a+12b+11=0$

Question 21: A ray of light passing through the point $(1,2)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $(5,3)$. Find the coordinates of $A$.

Answer:

From the figure above we can say that
The slope of line AC $(m)= \tan \theta$
Therefore,
$\tan \theta = \frac{3-0}{5-a} = \frac{3}{5-a} \ \ \ \ \ \ \ \ \ \ (i)$
Similarly,
The slope of line AB $(m') = \tan(180^\circ-\theta)$
Therefore,
$\tan(180^\circ-\theta) = \frac{2-0}{1-a}$
$-\tan\theta= \frac{2}{1-a}$
$\tan\theta= \frac{2}{a-1} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, from equation (i) and (ii) we will get
$\frac{3}{5-a} = \frac{2}{a-1}$
$\Rightarrow 3(a-1)= 2(5-a)$
$\Rightarrow 3a-3= 10-2a$
$\Rightarrow 5a=13$
$\Rightarrow a=\frac{13}{5}$
Therefore, the coordinates of $A$. is $\left ( \frac{13}{5},0 \right )$

Question 22: Prove that the product of the lengths of the perpendiculars drawn from the points $\small (\sqrt{a^2-b^2},0)$ and $\small (-\sqrt{a^2-b^2},0)$ to the line $\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ is $\small b^2$.

Answer:

Given equation id line is
$\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
We can rewrite it as
$xb\cos \theta +ya\sin \theta =ab$
Now, the distance of the line $xb\cos \theta +ya\sin \theta =ab$ from the point $\small (\sqrt{a^2-b^2},0)$ is given by
$d_1=\left | \frac{b\cos\theta.\sqrt{a^2-b^2}+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
Similarly,
The distance of the line $xb\cos \theta +ya\sin \theta =ab$ from the point $\small (-\sqrt{a^2-b^2},0)$ is given by
$d_2=\left | \frac{b\cos\theta.(-\sqrt{a^2-b^2})+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
$d_1.d_2 = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |.\times\left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
$=\left | \frac{-((b\cos\theta.\sqrt{a^2-b^2})^2-(ab)^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |$
$=\left | \frac{-b^2\cos^2\theta.(a^2-b^2)+a^2b^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |$
$=\left | \frac{-a^2b^2\cos^2\theta+b^4\cos^2\theta+a^2b^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2(\sin^2\theta+\cos^2\theta))}{b^2\cos^2\theta+a^2\sin^2\theta} \right | \ \ \ \ (\because \sin^2a+\cos^2a=1)$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2\sin^2\theta-a^2\cos^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{+b^2(b^2\cos^2\theta+a^2\sin^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=b^2$
Hence proved

Question 23: A person standing at the junction (crossing) of two straight paths represented by the equations $\small 2x-3y+4=0$ and $\small 3x+4y-5=0$ wants to reach the path whose equation is $\small 6x-7y+8=0$ in the least time. Find equation of the path that he should follow.

Answer:

point of intersection of lines $\small 2x-3y+4=0$ and $\small 3x+4y-5=0$ (junction) is $\left ( -\frac{1}{17},\frac{22}{17} \right )$
Now, person reaches to path $\small 6x-7y+8=0$ in least time when it follow the path perpendicular to it
Now,
Slope of line $\small 6x-7y+8=0$ is , $m'=\frac{6}{7}$
let the slope of line perpendicular to it is , m
Then,
$m= -\frac{1}{m}= -\frac{7}{6}$
Now, equation of line passing through point $\left ( -\frac{1}{17},\frac{22}{17} \right )$ and with slope $-\frac{7}{6}$ is
$\left ( y-\frac{22}{17} \right )= -\frac{7}{6}\left ( x-(-\frac{1}{17}) \right )$
$\Rightarrow 6(17y-22)=-7(17x+1)$
$\Rightarrow 102y-132=-119x-7$
$\Rightarrow 119x+102y=125$

Therefore, the required equation of line is $119x+102y=125$